The absolutely Koszul property of Veronese subrings and Segre products

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The absolutely Koszul property of Veronese subrings and Segre products Hop D. Nguyen Dipartimento di Matematica, Università di Genova, Via Dodecaneso 35, 16146 Genova, Italy

a r t i c l e

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Article history: Received 3 April 2017 Received in revised form 28 August 2017 Available online xxxx Communicated by G.G. Smith MSC: 13D02; 13D05

a b s t r a c t Absolutely Koszul algebras are a class of rings over which any finite graded module has a rational Poincaré series. We provide a criterion to detect non-absolutely Koszul rings. Combining the criterion with Macaulay2 computations, we identify large families of Veronese subrings and Segre products of polynomial rings which are not absolutely Koszul. In particular, we classify completely the absolutely Koszul algebras among Segre products of polynomial rings, when the base field has characteristic 0. © 2017 Elsevier B.V. All rights reserved.

1. Introduction Let k be a field, R be a standard graded k-algebra. Let M be a finitely generated graded R-module. An important problem in the theory of free resolutions concerns the rationality of the Poincaré series R PM (t) =

∞ 

βi (M )ti ∈ Q[[t]],

i=0

where βi (M ) = dimk TorR i (k, M ) denotes the i-th Betti number of M . While it is well-known that rationality does not hold in general, there are large classes of rings where every module has a rational Poincaré series, for example complete intersections [3], [18] and Golod rings [28] and [2, Theorem 5.3.2]. For a remarkable example, assume that char k = 0 and R = k[x1 , . . . , xn ]/I s where I is a non-zero proper homogeneous ideal of k[x1 , . . . , xn ] and s ≥ 2. Herzog and Huneke [21] proved recently that any such R is a Golod ring, in particular any finitely generated graded module over R has rational Poincaré series. Following Roos [33], we say that a standard graded k-algebra R is good if there exists a polynomial den(t) ∈ Z[t] such that for every finitely generated graded module M , R PM (t) · den(t) ∈ Z[t].

E-mail address: [email protected]. http://dx.doi.org/10.1016/j.jpaa.2017.09.017 0022-4049/© 2017 Elsevier B.V. All rights reserved.

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Otherwise we say R is bad. We say that R has the Backelin–Roos property if there exist a complete intersection Q and a Golod homomorphism Q → R (see [1,2,29] for more details on Golod homomorphisms). By a result attributed to Levin [5, Proposition 5.18] which was also proved by Backelin and Roos [7], if R has the Backelin–Roos property then it is good. Proving rationality of Poincaré series using the Backelin–Roos property is a powerful method; see, e.g., [4,5,19,34]. Note however that combining [27, pp. 291–292] and [26, Example 5.6], we have examples of (non-Koszul) good rings which do not have the Backelin–Roos property. A different method, among others, of establishing the rationality of Poincaré series comes from work of R Herzog and Iyengar [22]. They proved that PM (t) is a rational function with constant denominator, if M has finite linearity defect (see Section 2). Moreover, from [22, Theorem 5.9], if R is a Koszul algebra having the Backelin–Roos property, then any finitely generated module has finite linearity defect. Following the terminology in [24], we call R an absolutely Koszul algebra if every finitely generated graded R-module has finite linearity defect. Hence any absolutely Koszul algebra is good. So far, there was no example of a good Koszul algebra which is not absolutely Koszul, and no example of an absolutely Koszul algebra which does not have the Backelin–Roos property. The following basic question about absolutely Koszul rings was raised in [11, Section 4.2, Question 14]. Question 1.1. How do absolutely Koszul algebras behave with respect to algebra operations like taking Veronese subrings, Segre products or fiber products? The question is non-trivial since while Koszul algebras behave well with respect to standard operations [6], [8], it is not clear if the same thing happens to absolutely Koszul algebras. In fact, Question 1.1 has a negative answer for tensor product: For R = k[x, y]/(x, y)2 , R ⊗k R is not even good, not to mention absolutely Koszul [33, Theorem 2.4]. For fiber product, the answer to Question 1.1 is yes: A fiber product over k of two standard graded k-algebras is absolutely Koszul if and only if both factors are so ([12, Theorem 2.6]). The cases of Veronese subring and Segre product require more work, and so far only partial answers are available. Proposition 1.2 ([12, Corollary 5.4]). Assume char k = 0. Let k[x1 , . . . , xn ] be a polynomial ring in n variables and c ≥ 2 an integer. The c-th Veronese subring k[x1 , . . . , xn ](c) of k[x1 , . . . , xn ] is absolutely Koszul in the following cases: (1) c = 2 and n ≤ 6; (2) c ∈ {3, 4} and n ≤ 4; (3) c ≥ 5 and n ≤ 3. Proposition 1.3 ([12, Proposition 5.9]). Assume char k = 0 and 1 ≤ m ≤ n are integers. The Segre product Sm,n of the polynomial rings k[x1 , . . . , xm ] and k[y1 , . . . , yn ] is absolutely Koszul when (1) m ≤ 2, (2) m = 3 and n ≤ 5, (3) m = n = 4. Some evidence suggests that not all Veronese subrings and Segre products of polynomial rings are absolutely Koszul. It is proved in [12, Lemma 5.7] that for (n, c) equals either (7, 2), (5, 3), (5, 4), or (4, c), with c ≥ 5, k[x1 , . . . , xn ](c) does not have the Backelin–Roos property. The same thing happens for Sm,n if (m, n) equals either (3, 6) or (4, 5). Our study in this paper complements Propositions 1.2 and 1.3 by providing classes of Veronese rings and Segre products which are not absolutely Koszul.

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Theorem 1.4. Assume char k = 0. Let n ≥ 1 and c ≥ 2 be integers. Then the Veronese ring k[x1 , . . . , xn ](c) is not absolutely Koszul in the following cases: (1) c = 2 and n ≥ 7; (2) c ∈ {3, 4} and n ≥ 5; (3) c = 5 and n ≥ 4. Theorem 1.5. Assume char k = 0 and 1 ≤ m ≤ n be integers. The Segre product Sm,n of the polynomial rings k[x1 , . . . , xm ] and k[y1 , . . . , yn ] is not absolutely Koszul in the following cases (1) m = 3 and n ≥ 6, (2) m ≥ 4 and n ≥ 5. In particular, the only absolutely Koszul Sm,n are provided by Proposition 1.3. It is likely that our main results also hold true in other characteristics. But we do not know whether the Veronese ring k[x1 , . . . , xn ](c) where c ≥ 6 and n ≥ 4 is absolutely Koszul or not. In our opinion, studying further such Veronese rings is necessary for a complete answer to Question 1.1 and is an interesting challenge itself. Our proofs of the main results base on a criterion for non-absolutely Koszul rings; see Lemma 3.2 and Corollary 3.3. The idea of the criterion is simple. Assume that R is a standard graded k-algebra, and l1 is a linear form. Then the ideal M = (l1 ) has positive linearity defect if its first syzygy module, namely (0 : l1 )(−1) is not generated in degree 2, namely (0) : l1 is not generated by linear forms. Roughly speaking, Corollary 3.3 says that M has infinite linearity defect if, moreover, the second syzygy module of M has a direct summand which is isomorphic to a shifted copy of M : the non-linearity of the first syzygy module will “propagate” throughout the minimal free resolution of M . Corollary 3.3 is loosely spoken “opposite” to a criterion for absolutely Koszul rings via exact pairs of zero divisors due to Henriques and Şega [19]; see Remark 3.5. In order to use our criterion for non-absolutely Koszul rings for proving Theorems 1.4 and 1.5, we rely on extensive search with Macaulay2 [17]; the strategy of our search is described at the beginning of Section 4. Such costly computations are necessary to compensate for the non-constructive nature of the criterion, i.e. its silence on how to construct the linear form l1 . It would be interesting to seek for more conceptual proofs of the main results of this paper. The paper is organized as follows. After a background section, we present the afore-mentioned criterion for non-absolutely Koszul rings in Section 3. Applications of this criterion to the non-absolute Koszulness of certain Segre products and Veronese subrings are presented in Sections 4 and 5. We close the paper by discussing some open problems. 2. Background We assume familiarity with standard terminology and knowledge of commutative algebra, for which the books [9] and [14] serve as good references. 2.1. Linearity defect We recall the notion of linearity defect which Herzog and Iyengar introduced in [22], based in the notion of the linear part of a minimal free resolution. The linear part appeared in work of Herzog et al. [23, Section 5] and Eisenbud et al. [15]. Let (R, m, k) be a noetherian local ring, and M a finitely generated R-module.

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Denote grm R = ⊕i≥0 mi /mi+1 the associated graded ring of R with respect to the m-adic filtration. Let the minimal free resolution of M be ∂

∂

∂

∂

∂

F : ··· − → Fi − → Fi−1 − → ··· − → F1 − → F0 → 0. Since F is minimal, it admits a filtration · · · ⊆ F i F ⊆ F i−1 F ⊆ · · · ⊆ F 0 F = F as follows: F i F : · · · → Fj → Fj−1 → · · · → Fi → mFi−1 → · · · → mi−1 F1 → mi F0 → 0. The associated graded complex linR F =

 F iF F i+1 F i≥0

is called the linear part of F . Clearly linR F is a complex of graded grm R-modules with (linR F )i = (grm Fi )(−i). The linearity defect of M is ldR M = sup{i : Hi (linR F ) = 0}. By convention, if M = 0, we set ldR M = 0. When ldR M = 0, we say that M is a Koszul module. With appropriate changes, the above constructions also apply if R is a standard graded k-algebra with the graded maximal ideal m, and M a finitely generated graded R-module. The linearity defect can be characterized in terms of maps between Tor modules. Theorem 2.1 (Şega, [35, Theorem 2.2]). Let d ≥ 0 be an integer. The following are equivalent: (1) ldR M ≤ d, q+1 q q+1 (2) The map TorR , M ) → TorR → R/mq i (R/m i (R/m , M ) induced by the canonical surjection R/m is zero for all i > d and all q ≥ 0. Remark 2.2. A substantial simplification of Theorem 2.1 for R being a standard graded k-algebra and M a finitely generated graded R-module was discovered by Katthän [25, Theorem 1.3]. Using Theorem 2.1, we can prove the following useful statements. φ

→ P → N → 0 be a short Proposition 2.3 ([30, Corollary 2.10] and [31, Proposition 4.3]). Let 0 → M − exact sequence of finitely generated R-modules. (1) If P is free then ldR M = ldR N − 1 if ldR N ≥ 1 and ldR M = 0 if ldR N = 0. (2) Assume that TorR i (k, φ) = 0 for all i ≥ 0. Then there are inequalities: ldR M ≤ max{ldR P, ldR N − 1}, ldR P ≤ max{ldR M, ldR N }, ldR N ≤ max{ldR M + 1, ldR P }.

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2.2. Koszul algebras and regularity Let R be a standard graded k-algebra. Namely R is an N-graded commutative algebra with R0 = k such that R is generated by finitely many elements of degree 1 over R0 . Let M be a finitely generated graded R-module. The regularity of M over R is regR M = sup{j − i : TorR i (k, M )j = 0}. We say that M has a linear resolution over R if there exists d ∈ Z such that TorR i (k, M )j = 0 for all j = i + d. (We also say that M has a d-linear resolution in that case.) Modules with linear resolutions are typical examples of Koszul modules. We say that R is a Koszul algebra if the residue field k has a linear resolution over R. By Römer’s [32, Theorem 3.2.8], which extends Yanagawa’s [36, Proposition 4.9], if R is a Koszul algebra, then M is a Koszul module if and only if for each d ∈ Z, the submodule Md of M generated by elements of degree d has a d-linear resolution. A collection F of ideals generated by linear forms of R is a Koszul filtration [13] if the following conditions are satisfied: (1) (0), m ∈ F. (2) For every I = (0) in F, there exist an ideal J ∈ F properly contained in I such that I/J is a cyclic module and J : I ∈ F. If R has a Koszul filtration F, then any ideal I of F has a linear resolution over R. In particular, R is a Koszul algebra. It is useful to recall the following folkloric criterion for a module to have a linear resolution. Let R be a standard graded k-algebra, and M a finitely generated graded module. We say that M has linear quotients if there exist minimal homogeneous generators m1 , . . . , ms of M such that for all i = 0, 1, . . . , s − 1, the ideal (m1 , . . . , mi−1 ) : mi has a 1-linear resolution over R. Lemma 2.4. Let R be a standard graded k-algebra, and M a finitely generated graded module with linear quotients. Assume that M is generated by elements of the same degree d. Then M has a d-linear resolution. Proof. Assume that M is minimally generated by m1 , . . . , ms , all of degree d, and Ij = (m1 , . . . , mj−1 ) : mj has a 1-linear resolution for all j = 1, . . . , s. An easy induction shows that (m1 , . . . , mj ) has d-linear resolution for all j = 1, . . . , s. 2 2.3. Absolutely Koszul rings Let (R, m, k) be a noetherian local ring with the unique maximal ideal m, or a standard graded k-algebra with the graded maximal ideal m. Herzog and Iyengar [22, Proposition 1.8] proved that if M is a finitely R generated (graded) R-module such that ldR M < ∞, then PM (t) is a rational function with constant denominator (depending only on R). We say that R is absolutely Koszul if ldR M < ∞ for every finitely generated (graded) R-module M . If R is a graded absolutely Koszul algebra, then it is Koszul and good in the sense of Roos [33]. For the remainder of this section, we will restrict ourselves to the graded case. The following two results are useful to reduce the embedding dimension when proving that certain ring is not absolutely Koszul. The first was stated for local rings but its graded analogue is immediate.

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Theorem 2.5 ([30, Theorem 5.2]). Let R → S be a surjection of standard graded k-algebras such that S has a linear resolution as an R-module. Let N be a finitely generated graded S-module. Then there is an equality ldS N = ldR N . In particular, if R is an absolutely ring then so is S. θ

Recall that a homomorphism of standard graded k-algebras S − → R is called an algebra retract if there φ is a homomorphism of graded k-algebras R − → S such that φ ◦ θ = idS . θ

Lemma 2.6 (See [12, Proposition 2.3]). Let S − → R be an algebra retract of standard graded k-algebras. If R is absolutely Koszul then so is S. We say that R has the Backelin–Roos property if there exists a Golod map of standard graded k-algebras Q → R where the defining ideal of Q is generated by a regular sequence (see [1,29]). If R is Koszul, then by φ

[22, Proposition 5.8], R has the Backelin–Roos property if and only if there exists a map Q − → R, where Q is defined by a regular sequence of quadratic polynomials, and Ker φ has 2-linear resolution over Q. By [22, Theorem 5.9], if R is Koszul with the Backelin–Roos property, then R is absolutely Koszul. The Hilbert series of R is HSR (t) =

∞  (dimk Ri )ti ∈ Q[[t]]. i=0

By the Hilbert–Serre’s theorem, we can write HSR (t) =

hR (t) , (1 − t)dim R

where hR (t) ∈ Z[t] is called the h-polynomial of R. There is an obstruction on the Hilbert function of Koszul algebras with the Backelin–Roos property. Proposition 2.7 ([12, Proposition 3.12, Corollary 3.13]). Assume that R is a Koszul algebra with the Backelin–Roos property. Denote by codim R the codimension of R. Then the formal power series 1−

hR (−t) (1 − t)codim R

has only non-negative coefficients. Write hR (t) = (1 + t)s g(t), where s ≥ 0 and g(t) ∈ Z[t] is such that g(−1) = 0. If R is furthermore not a complete intersection, then g(−1) < 0. 3. A criterion for non-absolutely Koszul rings Let (R, m, k) be a noetherian local ring. Let M be a non-trivial finitely generated R-module and M1 , M2 be non-trivial submodules such that M = M1 +M2 . Following [16], we say that the decomposition M = M1 +M2 is a Betti splitting if for all i ≥ 0, there is an equality βi (M ) = βi (M1 ) + βi (M2 ) + βi−1 (M1 ∩ M2 ). Using the long exact sequence of Tor associated with the short exact sequence 0 → M1 ∩ M2 → M1 ⊕ M2 → M → 0, we can easily prove the following are equivalent: (1) M = M1 + M2 is a Betti splitting;

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R R R (2) For all i ≥ 0, the natural maps TorR i (k, M1 ∩ M2 ) → Tori (k, M1 ) and Tori (k, M1 ∩ M2 ) → Tori (k, M2 ) are zero.

In the graded case, we define Betti splittings with suitable modifications. We record the following simple lemma, which is useful to establish graded Betti splittings. φ

Lemma 3.1. Let R be a standard graded k-algebra. Let M − → N be morphism of finitely generated graded TorR (k,φ)

i R-modules. Assume that Mi = 0 for all i ≤ regR N . Then TorR −−− −−−→ TorR i (k, M ) − i (k, N ) is the zero map for all i ≥ 0.

R Proof. Denote d = regR N . For any j, consider the induced map TorR i (k, M )j → Tori (k, N )j . If j > i + d, R we have Tori (k, N )j = 0. If j ≤ i + d, since M has no generator of degree less than d + 1, we have TorR i (k, M )j = 0. The desired conclusion follows. 2

The following lemma is our main tool in proving that an algebra is not absolutely Koszul. Lemma 3.2. Let (R, m, k) be a noetherian local ring. Let p, q ≥ 1 be integers and φ1 : Rp → Rq , φ2 : Rq → Rp be non-zero maps of free R-modules such that φ1 ◦ φ2 = 0 and φ2 ◦ φ1 = 0. Assume that the following conditions are satisfied: (1) Both Im φ1 and Im φ2 are not Koszul modules; (2) There exist non-trivial submodules K1 ⊆ Ker φ2 , K2 ⊆ Ker φ1 satisfying the following two conditions: (a) There are Betti splittings Ker φ1 = Im φ2 + K2 and Ker φ2 = Im φ1 + K1 , (b) Each of the modules K1 ∩ Im φ1 , K2 ∩ Im φ2 is either zero or Koszul. Then ldR (Im φ1 ) = ldR (Im φ2 ) = ∞. In particular, R is not absolutely Koszul. Proof. Denote M = Im φ1 , N = Im φ2 . Since there is an exact sequence 0 −→ Ker φ1 −→ Rp −→ M −→ 0 and ldR M ≥ 1, by Proposition 2.3(1), ldR Ker φ1 = ldR M − 1. Consider the exact sequence 0 −→ K2 ∩ N −→ K2 ⊕ N −→ Ker φ1 −→ 0. R Since the decomposition Ker φ1 = K2 + N is a Betti splitting, the map TorR i (k, K2 ∩ N ) → Tori (k, K2 ⊕ N ) is zero for all i ≥ 0. Using Proposition 2.3(2), we get

max{ldR K2 , ldR N } ≤ max{ldR (K2 ∩ N ), ldR Ker φ1 } = ldR M − 1. The equality holds since ldR (K2 ∩ N ) = 0. Therefore ldR N ≤ ldR M − 1. Similarly, ldR M ≤ ldR N − 1. These inequalities cannot hold simultaneously unless ldR M = ldR N = ∞. 2 In practice, to prove that a Cohen–Macaulay algebra is not absolutely Koszul, we will usually pass to an artinian reduction using a regular sequence of linear forms, and use the following special case of (the graded analogue of) Lemma 3.2.

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Corollary 3.3. Let R be a standard graded k-algebra. Let l1 , l2 be non-zero linear forms such that l1 l2 = 0. Assume that there exist ideals (0) = K1 , K2 not generated by linear forms such that the following are satisfied: (1) (0) : l1 = K2 + (l2 ) and (0) : l2 = K1 + (l1 ); (2) The decompositions (0) : l1 = K2 + (l2 ) and (0) : l2 = K1 + (l1 ) are Betti splittings, e.g. K1 ∩ (l1 ) = K2 ∩ (l2 ) = (0); (3) Each of the modules K1 ∩ (l1 ) and K2 ∩ (l2 ) is either zero or Koszul. Then ldR (l1 ) = ldR (l2 ) = ∞ and hence R is not absolutely Koszul. Proof. If M is a finitely generated graded R-module, which is generated in a single degree d, then M ∼ = grm (M )(−d) as graded R-modules. Hence from [22, Proposition 1.5], ldR M = 0 if and only if M has a d-linear resolution over R. By the assumption (1), we conclude that ldR (l1 ) ≥ 1 and ldR (l2 ) ≥ 1. Now ·l

·l

1 2 using the graded analogue of Lemma 3.2 for the maps R(−1) −−→ R and R(−1) −−→ R we get ldR (l1 ) = ldR (l2 ) = ∞. 2

Remark 3.4. Roughly speaking, the hypotheses of Corollary 3.3 ensure that the homology of the linear part of the free resolution of (l1 ) is non-trivial at every homological degree. For simplicity, assume that K1 ∩ (l1 ) = K2 ∩ (l2 ) = (0). Let F be the minimal graded free resolution of (l1 ). Then as K2 ⊆ (0) : l1 , and K2 is not generated by linear forms, a minimal first syzygy of (l1 ) induces a non-zero element in H1 (linR F ). A similar argument works for (l2 ). Since (0) : l1 = K2 ⊕ (l2 ), the first syzygies of (l2 ) is a direct sum of the second syzygies of (l1 ). In particular, a minimal second syzygy of (l1 ) induces a non-zero element in H2 (linR F ). Moreover, the same thing happens for (l2 ). Repeating this argument, we see that the linear part of F has non-zero homology at every homological degree. Remark 3.5. Henriques and Şega introduced in [19] the notion of an exact pair of zero divisors. Assume that R is a standard graded k-algebra, and x, y are homogeneous elements in R such that xy = 0. We say that x and y form an exact pair of zero divisors if 0 : x = (y) and 0 : y = (x). If x and y form an exact pair of zero divisors and both are linear forms, then ldR (x) = 0 since (x) has the linear resolution ·x

·y

·x

·y

· · · −→ R(−4) −→ R(−3) −→ R(−2) −→ R(−1) → (x) → 0. Similarly ldR (y) = 0. One of the main results of [19], Theorem 3.3, yields a criterion for absolutely Koszul rings using exact pairs of zero divisors. The hypotheses and conclusion of Corollary 3.3 are in some sense “opposite” to that of [19, Theorem 3.3]. Example 3.6. Let R = k[x, y, z, u]/(x2 , xy, y 2 , z 2 , zu, u2 ). This was considered by Roos, who proved that R is a bad Koszul algebra, and in particular not absolutely Koszul [33, Theorem 2.4]. Corollary 3.3 yields a direct proof of the non-absolute Koszulness of R. The study of this example in fact motivated the non-absolute Koszulness criterion and the main results of this paper. Denote l1 = x − z, l2 = x + z. Then by straightforward calculations, we have (0) : l1 = (yu) + (l2 ), (0) : l2 = (yu) + (l1 ), (yu) ∩ (l1 ) = (yu) ∩ (l2 ) = (0). Hence the conditions of Corollary 3.3 are fulfilled. In particular, ldR (x − z) = ldR (x + z) = ∞ and R is not absolutely Koszul.

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Example 3.7. The following construction is taken from [10, Example 3.8]. Let Q = k[x, y, z, u] and R = Q/I where I = (x2 + yz, xy − yu, xz, xu, y 2 ). Let φ be the map R(−1)2 = Re1 ⊕ Re2 → R2 given by the matrix 

−x y z x

 .

We claim that ldR Im φ = ∞, whence R is not absolutely Koszul. First R is Koszul since R admits the following Koszul filtration (where by abuse of notation, we use f to denote the residue class of f ): {(0), (x), (z), (x, z), (x, y), (z, u), (x, z, u), (x, y, z, u)} . In detail, by Macaulay2 [17], the following equalities hold if char k ≤ 4000 (and probably in all characteristics): (0) : (x) = (z, u), (0) : (z) = (x), (z) : (x, z) = (x, z, u), (x) : (x, y) = (x, y, z, u), (z) : (z, u) = (x, z), (z, u) : (x, z, u) = (x, y, z, u), (x, z, u) : (x, y, z, u) = (x, y, z, u). Denote M = Im φ. Now Ker φ equals the image of the map R(−2)2 ⊕ R(−3) → R(−1)2 = Re1 ⊕ Re2 given by the matrix   −x y 0 . z x u2 In particular φ2 = 0. We can identify (Im φ)(−1) = M (−1) with (−xe1 + ze2 , ye1 + xe2 ) and letting K = (u2 e2 ), we get a decomposition of the first syzygy of M : ΩR 1 (M ) = Ker φ = M (−1) + K. Note that K ∼ = (R/(x, y))(−3), so K has a 3-linear resolution. In the same way, let L = M (−1) ∩ K, then L = (zu2 e2 ) ∼ = (R/(x, y))(−4) has a 4-linear resolution. We will show that regR M = 2. Denote ti (M ) = sup{j : TorR i (k, M )j = 0}, then regR M = supi≥0 {ti (M ) − i}. Since t1 (M ) = 3, regR M ≥ 2, thus it suffices to show that ti (M ) ≤ i + 2 for all i ≥ 0. Induct on i; the cases i = 0, 1 are clear. Consider the exact sequence 0 −→ L −→ M (−1) ⊕ K −→ ΩR 1 (M ) −→ 0. For i ≥ 2, we have ti (M ) = ti−1 (ΩR 1 (M )) ≤ max{ti−1 (M ) + 1, ti−1 (K), ti−2 (L)} = max{ti−1 (M ) + 1, i + 2} = i + 2.

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In the display, the second equality holds since K has a 3-linear and L has a 4-linear resolution. The last equality holds by the induction hypothesis. Therefore regR M = 2. Since M is generated in degree 1 and R is Koszul, we conclude that M is not Koszul. Moreover, by Lemma 3.1, the decomposition ΩR 1 (M ) = M (−1) + K is a Betti splitting. Hence by the graded analogue of Lemma 3.2, ldR M = ∞. Notably, while R is not an integral domain (as its socle contains yz = 0), there exists no pair l1 and l2 of linear forms of R that satisfy the conditions of Corollary 3.3. We give a sketch of the argument here. Assume the contrary, that there exist such a pair of linear forms l1 , l2 . Denote by f, g the linear forms of Q = k[x, y, z, u] representing the linear forms l1 , l2 of R. Working in Q, since f g ∈ I ⊆ (x, y), either f or g belongs to (x, y). We can assume that f ∈ (x, y). By elementary considerations, up to scaling, only the following cases are possible, where a, b, α ∈ k, (a, b) = (0, 0), α = 0: (1) f = y, g = a(x − u) + by, (2) f = x, g = az + bu,   (3) f = x + αy, g = a x − αy + (1/α)z + b(αy − u). Now we return to R. In Case (2), (0) : l2 = (x) is generated by a linear form, a contradiction. In Case (3), (0) : l2 = (l1 ), again a contradiction. In Case (1), we must have a = 0, otherwise (0) : l2 = (l1 ). Thus it remains to consider the case (l1 ) = (l2 ) = (y). We have (0) : y = (y, x − u, z 2 ). By assumption, there exists a submodule K ⊆ (y, x − u, z 2 ) = U such that U = (y) + K is a Betti splitting. Since R R R R R TorR i (k, (y) ∩ K) → Tori (k, (y)) ⊕ Tori (k, K) → Tori (k, U ) is exact, the map Tori (k, (y)) → Tori (k, U ) is injective for all i ≥ 0. In particular, for all i, βi (U ) = βi ((y)) + βi (U/(y)). But β1 (U ) = 7 = β1 ((y)) + β1 (U/(y)) = 3 + 5. This contradiction finishes the proof. This example shows that the criterion of Lemma 3.2 is stronger than that of Corollary 3.3. Nevertheless for the main applications of this paper, it suffices to invoke Corollary 3.3. 4. Segre products In the current and next section, we prove the main Theorems 1.5 and 1.4. Our general strategy in both cases is as follows. Suppose that R is a Koszul Cohen-Macaulay k-algebra which is not a complete intersection such that its h-polynomial hR (t) satisfies hR (−1) > 0, and we want to show that R is not absolutely Koszul. (This is the case for the rings considered in Sections 4 and 5.) First pass to an infinite field extension of k if necessary, and let S = R/(l1 , . . . , ld ), where d = dim R, and l1 , . . . , ld is a regular sequence of linear forms of R. By Theorem 2.5, it suffices to prove that S is not absolutely Koszul. Note that S still satisfies hS (−1) > 0 since the h-polynomial does not change after an artinian reduction. To simplify the problem, we find an ideal I generated by linear forms of S such that S/I has linear resolution over S, and hS/I (−1) > 0. We can ensure regS (S/I) = 0 by constructing a Koszul filtration of S containing I. Since hS/I (−1) > 0, Proposition 2.7 gives that S/I does not have the Backelin–Roos property. Then we construct suitable linear forms l1 and l2 in order to apply Corollary 3.3 to S/I and deduce that it is not absolutely Koszul. Then thanks to Theorem 2.5, R is also not absolutely Koszul, and we are done. The most tricky part in this strategy is the search for l1 and l2 , for which we depend heavily on computer assistance. The relevant Segre products have lower embedding dimensions, so we treat them before the Veronese rings. Recall that if A and B are standard graded k-algebras, then their Segre product A ∗B is the subalgebra of A ⊗k B with the graded component (A ∗ B)i = Ai ⊗k Bi for all i ≥ 0.

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In the following, for 1 ≤ m ≤ n, let Sm,n denote the Segre product k[x1 , . . . , xm ] ∗ k[y1 , . . . , yn ]. The Hilbert function of Sm,n is given by 

  n+i−1 , n−1

m+i−1 m−1

HF(Sm,n , i) = dimk Ai dimk Bi =

for i ≥ 1.

In particular, the h-polynomial hSm,n (t) of Sm,n is

hSm,n (t) =

m−1  i=0

  n−1 i t i

m−1 i

and its embedding dimension and Krull dimension are embdim(Sm,n ) = mn and dim Sm,n = m + n − 1, respectively. Proposition 4.1. Assume that char k = 0. Then the Segre product of k[x1 , x2 , x3 ] and k[y1 , y2 , . . . , y6 ] is not absolutely Koszul. Proof. The ring R = S3,6 is defined by the 2-minors of the generic matrix 

z11 z21 z31

z12 z22 z32

z13 z23 z33

z14 z24 z34

z15 z25 z35

z16 z26 z36



Denote U = R/J, where J = (z11 , z12 − z21 , z13 − z22 − z31 , z14 − z23 − z32 , z15 − z24 − z33 , z16 − z25 − z34 , z26 − z35 , z36 ) is generated by a regular sequence of length 8 = dim R. Then U is a quotient ring B/H of B = k[z21 , z22 , . . . , z34 , z35 ]. The ring U has Hilbert series 1 + 10t + 10t2 . Rename the variables of B in the dictionary order: a1 = z21 , a2 = z22 , . . . , a10 = z35 . Then B = k[a1 , . . . , a10 ]. In particular, H is the ideal of 2-minors of the matrix 

0 a1 a6

a1 a2 a7

a 2 + a6 a3 a8

a3 + a7 a4 a9

a4 + a8 a5 a10

a5 + a9 a10 0



Denote l1 = a1 − a3 + a5 − a7 , l2 = a2 + a6 − a8 − a10 . Then we have (0) : l1 = (l2 , a3 a9 ), (0) : l2 = (l1 , a4 a10 ), (l2 ) ∩ (a3 a9 ) = (l1 ) ∩ (a4 a10 ) = (0). By Corollary 3.3, U is not absolutely Koszul, and as regR U = 0, neither is R by Theorem 2.5. 2 Proposition 4.2. Assume that char k = 0. The Segre product of k[x1 , x2 , x3 , x4 ] and k[y1 , y2 , y3 , y4 , y5 ] is not absolutely Koszul.

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Proof. The ring R = S4,5 is defined by the 2-minors of the generic matrix ⎛

z11 ⎜ z21 ⎝z 31 z41

z12 z22 z32 z42

z13 z23 z33 z43

z14 z24 z34 z44

⎞ z15 z25 ⎟ z35 ⎠ z45

Denote U = R/J, where J = (z11 , z12 − z21 , z13 − z22 − z31 , z14 − z23 − z32 − z41 , z15 − z24 − z33 − z42 , z25 − z34 − z43 , z35 − z44 , z45 ) is generated by a regular sequence of length 8 = dim R. Then U is a quotient ring B/H of B = k[z21 , z22 , . . . , z43 , z44 ]. The ring U has Hilbert series 1 + 12t + 18t2 + 4t3 . Rename the variables of B in the dictionary order: a1 = z21 , a2 = z22 , . . . , a12 = z44 . Then B = k[a1 , . . . , a12 ]. In particular, H is the ideal of 2-minors of the matrix ⎛

0 ⎜ a1 ⎝a 5 a9

a1 a2 a6 a10

a2 + a5 a3 a7 a11

a3 + a6 + a9 a4 a8 a12

⎞ a4 + a7 + a10 a8 + a11 ⎟ ⎠ a12 0

Step 1: Let m be the graded maximal ideal of U . We claim that there exists a Koszul filtration F of U which contains the following ideals: (0), (a12 ), m. In detail, let F be the collection of ideals (0), I1 = (a12 ), I2 = (a12 , a11 ), I3 = (a12 , a11 , a10 ), I4 = (a12 , a11 , a10 , a9 ), I5 = (a12 , a11 , . . . , a8 ), I6 = (a12 , a11 , . . . , a8 , a7 + a4 ), I7 = (a12 , a11 , . . . , a7 , a4 ), I8 = (a12 , a11 , . . . , a7 , a4 , a6 + a3 ), I9 = (a12 , a11 , . . . , a7 , a6 , a4 , a3 ), I10 = (a12 , a11 , . . . , a6 , a5 + a2 , a4 , a3 ), I11 = (a12 , a11 , . . . , a3 , a2 ), I12 = (a12 , a11 , . . . , a3 ), I13 = (a12 , a11 , . . . , a6 , a4 ), I14 = (a12 , a11 , . . . , a4 ), m. Then F is a Koszul filtration since we have the following identities (0) : I1 = I6 ,

I1 : I2 = I8 ,

I2 : I3 = I10 ,

I3 : I4 = m,

I4 : I5 = I14 ,

I5 : I6 = m,

I6 : I7 = I12 ,

I7 : I8 = m,

I8 : I9 = I11 ,

I9 : I10 = m,

I10 : I11 = m,

I9 : I12 = m,

I7 : I13 = I11 ,

I13 : I14 = m,

I11 : m = m.

Step 2: Denote W = U/(a12 ). Note that the Hilbert series of W is 1 + 11t + 12t2 + t3 . Let n be the graded maximal ideal of W , then n3 = (a1 a6 a11 ) = 0 and n4 = 0. As we have seen, regU W = 0 = regR U . If R was absolutely Koszul, we conclude by Theorem 2.5 that so is W . On the other hand, letting l1 = a2 +a4 −a6 +a10 , l2 = a2 + 2a3 + 2a5 + 3a6 − a8 + 6a9 − a10 , K = (a6 a11 ), we have

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(0) : l1 = K + (l2 ), (0) : l2 = K + (l1 ), K ∩ (l1 ) = K ∩ (l2 ) = (a1 a6 a11 ) = n3 .

(4.1)

We claim that: (1) regW K = 2, (2) regW (l1 ) = regW (l2 ) = 2. For (1): Since F contains the ideals (a12 ) and I11 = (a12 , a11 , . . . , a3 , a2 ), and (a12 ) is the minimal non-zero element of F, there is an induced Koszul filtration of W which contains (a11 , a10 , . . . , a3 , a2 ) = (0) : (a6 a11 ). Hence regW (a6 a11 ) = 2. For (2): Clearly regR (l1 ) ≥ 2 and regR (l2 ) ≥ 2. For the reverse inequalities, we prove by induction on i that ti ((l1 )) ≤ i + 2 and ti ((l2 )) ≤ i + 2 for all i ≥ 0. The cases i = 0, 1 are clear. Assume that i ≥ 2. Denote L = n3 ∼ = (W/n)(−3), which has a 3-linear resolution. Consider the exact sequence 0 −→ L −→ K ⊕ (l2 ) −→ (0) : l1 −→ 0. The module ((0) : l1 )(−1) being the first syzygy of (l1 ), we obtain ti ((l1 )) = ti−1 ((0) : l1 ) + 1 ≤ max{ti−1 (K), ti−1 ((l2 )), ti−2 (L)} + 1 = max{i + 2, ti−1 ((l2 )) + 1} = i + 2. In the chain, the second equality holds since K has a 2-linear resolution, and L has a 3-linear resolution. The last equality holds by induction hypothesis. Similarly, ti ((l2 )) ≤ i + 2, and we finish the proof of (2). From (1), (2) and (4.1), the decompositions (0) : l1 = K + (l2 ) and (0) : l2 = K + (l1 ) are Betti splittings by an application of Lemma 3.1. By Corollary 3.3, W is not absolutely Koszul. This contradiction implies that S4,5 is not absolutely Koszul. 2 From Propositions 4.1, 4.2, we easily deduce Theorem 1.5. Proof of Theorem 1.5. Apply Lemma 2.6 and the fact that if m ≤ m , n ≤ n then there is an algebra retract Sm,n → Sm ,n . 2 5. Veronese rings In this section, we prove Theorem 1.4. Let S be a standard graded k-algebra. Denote S (c) = ⊕∞ i=0 Sic the c-th Veronese subring of S, whose grading is given by deg Sic = i for all i. For n, c ≥ 1, denote by Vn,c the ring k[x1 , . . . , xn ](c) . The Hilbert function of Vn,c = k[x1 , . . . , xn ](c) is given by  HF(Vn,c , i) = In particular, the h-polynomial of Vn,c is

 n + ic − 1 , n−1

for all i ≥ 1.

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hVn,c (t) =

n−1  i=0

i

hi t ,

where hi =

i  j=0

 i−j

(−1)

  n − 1 + jc n . n−1 i−j

  The embedding dimension and Krull dimension of Vn,c are n+c−1 and n, respectively. c Let R be a standard graded k-algebra with the graded maximal ideal m. Assume that the ideal m is minimally generated by the linear forms a1 , . . . , an . We say that R is strongly Koszul (with respect to the sequence a1 , . . . , an ), if for every 1 ≤ r ≤ n and every sequence of pairwise distinct elements i1 , i2 , . . . , ir of [n] = {1, . . . , n}, the ideal (ai1 , . . . , air−1 ) : air is generated by a subset of {a1 , . . . , an }. If R is strongly Koszul with respect to the sequence a1 , . . . , an , then clearly the collection of ideals generated by subsets of {a1 , . . . , an } forms a Koszul filtration for R. Furthermore, for any sequence of pairwise distinct elements i1 , . . . , ir in [n], the ring R/(ai1 , . . . , air ) is also strongly Koszul. It is known that any Veronese subring of a polynomial ring is strongly Koszul ([20, Proposition 2.3]). Recall the statement of Theorem 1.4. Theorem 5.1. Assume char k = 0. Then Vn,c is not absolutely Koszul in the following cases: (1) c = 2 and n ≥ 7; (2) c ∈ {3, 4} and n ≥ 5; (3) c = 5 and n ≥ 4. Proof. If m ≤ n then there is an algebra retract Vm,c → Vn,c . Therefore from Lemma 2.6, we only need to show that Vn,c is not absolutely Koszul if (n, c) belongs to the set {(7, 2), (5, 3), (5, 4), (4, 5)}. Denote S = k[x1 , . . . , xn ] and R = Vn,c . Then R is generated as an algebra by the standard k-basis of Sc . The   vector space dimension of Sc is N = n+c−1 . Order the standard k-basis of Sc in the degree revlex order as c c c c c m1 < · · · < mN , so that m1 = xcn , m2 = xc−1 n xn−1 , . . . , mN = x1 . Clearly xn , xn−1 , . . . , x1 is a maximal regc c ular sequence of elements of degree 1 of R, so the artinian reduction U = R/(xn , xn−1 , . . . , xc1 ) of R has embedding dimension N −n. We have a surjection with the source a polynomial ring B = k[a1 , . . . , aN −n ]  U . One can obtain U with the following Macaulay2 function artinVeroneseSubring. Input: Integers n, c ≥ 1. Output: The artinian reduction U = Vn,c /(xcn , . . . , xc1 ) of Vn,c . The variables of the ambient ring of U are ordered as in the description above. artinVeroneseSubring=(n,c)->( toricR=QQ[x_1..x_n]; maxtoricR=ideal vars toricR; dsc=flatten entries mingens maxtoricR^c; dsg={}; for i from 1 to n do dsg=append(dsg,x_(n+1-i)^c); lengthdsc=length dsc; S=QQ[y_1..y_lengthdsc]; axavero=map(toricR,S,dsc); veroId=trim preimage(axavero,ideal(dsg)); U1=prune(S/veroId); Am=ambient U1; numvar=length flatten entries vars U1; newring=QQ[a_1..a_numvar];

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axanew=map(newring,Am,{a_1..a_numvar}); artinVeroId=trim axanew(ideal U1); newring/artinVeroId) Case 1: (n, c) = (7, 2). In this case N = 28, B = k[a1 , . . . , a21 ]. The Hilbert series of U is 1+21t+35t2 +7t3 . Denote W = U/(a21 ), n the graded maximal ideal of W . The Hilbert series hW (t) = 1 + 20t + 25t2 + 2t3 of W satisfies hW (−1) > 0, hence it at least does not have the Backelin–Roos property by Proposition 2.7. Let n be the graded maximal ideal of W , then n3 = (a1 a12 a19 , a1 a12 a20 ). Since R is a strongly Koszul ring, so is W , and regR W = 0. Denote l1 = a1 + a2 − a8 + a12 + a20 , l2 = a1 + a2 − a8 + a12 − a20 and K = (a2 + a12 , a3 , a7 , a8 − a12 )a19 . Then (0) : l1 = K + (l2 ), (0) : l2 = K + (l1 ), K ∩ (l1 ) = K ∩ (l2 ) = (a1 a12 a19 ).

(5.1)

We claim that (1) regW K = 2, (2) regW (l1 ) = regW (l2 ) = 2. For (1): K has linear quotients since W is strongly Koszul and (0) : (a3 a19 ) = (ai : i ∈ [20] \ 7), (a3 a19 ) : (a7 a19 ) = n, (a3 a19 , a7 a19 ) : ((a2 + a12 )a19 ) = n, (a3 a19 , a7 a19 , (a2 + a12 )a19 ) : ((a8 − a12 )a19 ) = n. By Lemma 2.4, regW K = 2. For (2): Denote L = K ∩ (l1 ) = (a1 a12 a19 ) ∼ = (W/n)(−3). Then L has a 3-linear resolution. Similarly to Step 2 in the proof of Proposition 4.2, we get regR (l1 ) = regR (l2 ) = 2. Now from (1), (2) and (5.1), it follows that the decompositions (0) : l1 = (K, l2 ) and (0) : l2 = (K, l1 ) are Betti splittings by an application of Lemma 3.1. Hence by Corollary 3.3, W is not absolutely Koszul. Since regR W = 0, by Theorem 2.5, R = V7,2 is not absolutely Koszul. Case 2: (n, c) = (5, 3). In this case N = 35, B = k[a1 , . . . , a30 ]. The Hilbert series of U is 1+30t+45t2 +5t3 . Denote W = U/(a30 ); its Hilbert series is 1 + 29t + 32t2 + t3 . Denote l1 = a1 + a2 − a5 + a24 + a29 , l2 = a1 + a2 − a5 − a24 − a29 and K = (a5 a28 , a6 a28 , a15 a28 ). We have (0) : l1 = K + (l2 ), (0) : l2 = K + (l1 ), K ∩ (l1 ) = K ∩ (l2 ) = (0). By Corollary 3.3, W is not absolutely Koszul. Since R is strongly Koszul, regR W = 0. Hence by Theorem 2.5, R = V5,3 is not absolutely Koszul.

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Case 3: (n, c) = (5, 4). In this case N = 70, B = k[a1 , . . . , a65 ]. The Hilbert series of U is 1 + 65t + 155t2 + 35t3 . Denote W = U/(a60 , a61 , . . . , a65 ); its Hilbert series is 1 + 59t + 63t2 + t3 . Denote l1 = a1 + a5 − a16 + a54 + a59 , l2 = a1 + a5 − a16 − a54 − a59 and K = (a2 a58 , a6 a58 , a16 a58 , a35 a58 ). We have (0) : l1 = K + (l2 ), (0) : l2 = K + (l1 ), K ∩ (l1 ) = K ∩ (l2 ) = (0). Hence by Corollary 3.3, ldW (l1 ) = ldW (l2 ) = ∞ and W is not absolutely Koszul. Since R is strongly Koszul, regR W = 0. Hence by Theorem 2.5, R = V5,4 is not absolutely Koszul. Case 4: (n, c) = (4, 5). In this case N = 56, B = k[a1 , . . . , a52 ]. The Hilbert series of U is 1 +52t +68t2 +4t3 . Denote W = U/(a52 ); its Hilbert series is 1 + 51t + 52t2 + t3 . Denote l1 = a1 + a2 − a10 + a45 + a51 , l2 = a1 + a2 − a10 − a45 − a51 and K = (a10 a48 ). Then there are equalities: (0) : l1 = K + (l2 ), (0) : l2 = K + (l1 ), K ∩ (l1 ) = K ∩ (l2 ) = (0). Hence thanks to Corollary 3.3, W is not absolutely Koszul. Since R is strongly Koszul, regR W = 0. Hence by Theorem 2.5, R = V4,5 is not absolutely Koszul. The proof is concluded. 2 Tempted by the theoretical and experimental results available thus far, we ask: Question 5.2. Is it true that the ring k[x1 , x2 , x3 , x4 ](c) is not absolutely Koszul for all c ≥ 6, when char k = 0? If this is true, then by Lemma 2.6, all the absolutely Koszul Veronese rings k[x1 , . . . , xn ](c) are described by Proposition 1.2. While many Veronese subrings and Segre products are not absolutely Koszul, it is not clear which of them are good. Hence we ask: Question 5.3. Are Veronese subrings and Segre products of polynomial rings good in the sense of Roos? Acknowledgements The author is a Marie Curie fellow of the Istituto Nazionale di Alta Matematica. He would like to thank Aldo Conca, Jan-Erik Roos and Marilina Rossi for inspiring discussions. He is also grateful to the anonymous referee for useful suggestions on the presentation. References [1] L.L. Avramov, Golod homomorphisms, in: Algebra, Algebraic Topology, and Their Interactions, Stockholm, 1983, in: Lect. Notes Math., vol. 1183, Springer, Berlin, 1986, pp. 56–78. [2] L.L. Avramov, Infinite free resolution, in: Six Lectures on Commutative Algebra, Bellaterra, 1996, in: Prog. Math., vol. 166, Birkhäuser, 1998, pp. 1–118. [3] L.L. Avramov, V.N. Gasharov, I. Peeva, Complete intersection dimension, Publ. Math. IHES 86 (1997) 67–114. [4] L.L. Avramov, S.B. Iyengar, L.M. Şega, Free resolutions over short local rings, J. Lond. Math. Soc. 78 (2) (2008) 459–476. [5] L.L. Avramov, A. Kustin, M. Miller, Poincaré series of modules over local rings of small embedding codepth or small linking number, J. Algebra 118 (1988) 162–204.

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